Problem
NT-B2-M10-P002 Number of Divisors
#2
★★☆☆☆ Level 2 of 5
Find all natural numbers \(n=2^a3^b\) that have exactly \(18\) positive divisors.
You need \((a+1)(b+1)=18\).
Since \(\tau(n)=(a+1)(b+1)\), the pairs are \((a+1,b+1)=(1,18),(2,9),(3,6),(6,3),(9,2),(18,1)\). Because \(a,b\ge 0\) are allowed in \(2^a3^b\), all pairs give solutions: \(3^{17}\), \(2\cdot 3^8\), \(2^2\cdot 3^5\), \(2^5\cdot 3^2\), \(2^8\cdot 3\), \(2^{17}\).
Clarify whether exponent \(0\) is allowed; here it is.